Chapter 8 Application Of Second-order Differential .
Applied Engineering Analysis- slides for class teaching*Chapter 8Application of Second-order Differential Equationsin Mechanical Engineering Analysis* Based on the book of “Applied EngineeringAnalysis”, by Tai-Ran Hsu, published byJohn Wiley & Sons, 2018 (ISBN 9781119071204)(Chapter 8 second order DEs) Tai-Ran Hsu1
Chapter Learning Objectives Refresh the solution methods for typical second-order homogeneous and nonhomogeneous differential equations learned in previous math courses, Learn to derive homogeneous second-order differential equations for freevibration analysis of simple mass-spring system with and without dampingeffects, Learn to derive nonhomogeneous second-order differential equations forforced vibration analysis of simple mass-spring systems, Learn to use the solution of second-order nonhomogeneous differentialequations to illustrate the resonant vibration of simple mass-spring systemsand estimate the time for the rupture of the system under in resonant vibration, Learn to use the second order nonhomogeneous differential equation to predictthe amplitudes of the vibrating mass in the situation of near-resonant vibrationand the physical consequences to the mass-spring systems, and Learn the concept of modal analysis of machines and structures and theconsequence of structural failure under the resonant and near-resonantvibration modes.2
Review Solution Method of SecondOrder, Homogeneous OrdinaryDifferential EquationsWe will review the techniques available for solvingtypical second order differential equations at thebeginning of this chapter.The solution methods presented in the subsequentsections are generic and effective for engineeringanalysis.3
8.2 Typicalform of second-order homogeneous differential equations (p.243)d 2u ( x )du ( x) a bu ( x) 0dx 2dx(8.1)where a and b are constantsThe solution of Equation (8.1) u(x) may be obtained by ASSUMING:(8.2)u(x) emxin which m is a constant to be determined by the following procedure:If the assumed solution u(x) in Equation (8.2) is a valid solution, it must SATISFYEquation (8.1). That is:d 2 e mxd e mxmx(a) a be 02 Because:2 mmxd edx 2 dxdx2emx d e mx me mxdxandSubstitution of the above expressions into Equation (a) will lead to: m 2 e mx a m e mx b e mx 0Because emx in the expression cannot be zero (why?), we thus have:m2 am b 0(8.3)Equation (8.3) is a quadratic equation with unknown “m”, and its 2 solutions for m are from:4
m1 a 1 2 a 4b2 2andm2 a 1 2 a 4b2 2(8.4)This leads to the following two possible solutions for the function u(x) in Equation (8.1):u x c1 e m1x c2 e m2 x(8.5)where c1 and c2 are the TWO arbitrary constants to be determined by TWO specifiedconditions, and m1 and m2 are expressed in Equation (8.4)Because the constant coefficients a and b in Equation (8.1) are given in the differentialequation, the values these constants a, b will result in significantly different forms in thesolution as shown in Equation (8.5) due to the “square root” parts in the expression ofm1 and m2 in Equation (8.4). Because square root of negative numbers will lead to a complexnumber in the solution of the differential equation, which requires a special way of expressing it.We thus need to look into the following 3 possible cases involving relative magnitudes of thetwo coefficients a and b in Equation (8.1).5
Case 1.a2 – 4b 0:In such case, we realize that both m1 and m2 are real numbers. The solution of theaxEquation (8.1) is: 22u ( x ) e 2 c1e a 4b x / 2 c2 e a 4b x / 2 (8.6) 2Case 2.a - 4b 0:As described earlier, both these roots become complex numbers involving real and imaginaryparts. The substitution of the m1 and m2 into Equation (8.5) will lead to the following:u ( x) e ax2ix c1 e 2 4b a 2 c2e ix4b a 22 (8.7)in which, i 1 . The complex form of the solution in Equation (8.7) is not always easilycomprehended and manipulative in engineering analyses, a more commonly used forminvolving trigonometric functionsare used instead:ax 1 1 (8.8)u ( x ) e 2 A Sin 4b a 2 x B Cos 4b a 2 x 2 2 where A and B are arbitrary constants to be determined by given conditions.The expression in Equation (8.8) may be derived from Equation (8.7) using the Biot relationthat has the form: e i Cos i Sin For a special case with coefficient a 0 and b is a negative number, the solution ofEquation (8.1) becomes:(8.9)u x c cosh 2 b x c sinh 2 b x12where c1 and c2 are arbitrary constants to be determined by given conditions.6
Case 3.a2 - 4b 0:Recall Equation (8.4):a 1 2m1 a 4b2 2andm2 a 1 2 a 4b2 2(b)The condition a2 – 4b 0 will thus lead to a situation of: m1 m2 - a/2Substituting these m1 and m2 into Equation (8.5) will result in:u ( x) (c1 c2 ) ea x2or u1 ( x) cea x2with only ONE term with ONE constant in the solution, which cannot be a complete solutionfor a 2nd order differential equation in Equation (8.1).We will have to find the “missing” solution of u(x) for a second-order differential equation inEquation (8.1) by following the procedure:. Let us try the following additional assumed form of the solution u(x) :u2(x) V(x) emx(8.10)where V(x) is an assumed function of x, and it needs to be determinedThe assumed second solution in Equation (8.10) must satisfy Equation (8.1)7
2du ( x)du ( x)The Differential equation: a bu ( x) 02dxdx(8.1)The assumed second solution to Equation (8.1) is: u2(x) V(x) emx , which leads to thefollowing equality: d 2 V x e mxd V x e mxmx a bVxe 02dxdxOne would find:and (c) d V x e mxdV x mV x e mx e mxdxdx 2d 2 V x e mx mxmx dV x mx dV x mx d V x me e m mV x e e2 dxdx dxdx 2 After substituting the above expressions into Equation (c), we will get: dV ( x)d 2V ( x)2 (2 ) m am b V ( x) 0ma2dxdx(8.11)Since m2 am b 0 in Equation (8.3), and m m1 m2 - a/2 in Equation (b),so both the 2nd and 3rd term in Equation (8.11) drop out. We thus only have the first termTo consider in the following special form of a 2nd order differential equation:d 2V ( x) 0dx 2The solution of the above differential equation is: V(x) x after 2 sequential integrations8
The solution V(x) x leads to the missing second solution of the differential equation inEquation (8.1)d 2u ( x )du ( x) a bu ( x) 0dx 2dx(8.1)in Case 3 with a2 – 4b 0 as:u2 x V ( x)e mx xe mx xe ax2The general solution of Equation (8.1) with a2-4b 0 thus becomes:u(x) u1(x) u2(x)oru ( x) c1 e ax2 c2 x e ax2 c1 c2 x e ax2(8.12)where the two arbitrary constants c1 and c2 are determined by the two given conditionswith Equation (8.1).9
Summary on Solutions of 2nd Order Homogeneous Differential Equationsd 2u ( x )du ( x) a bu ( x) 0dx 2dxThe equation:(8.1)with TWO given conditionsThe solutions:Case 1: a2 – 4b 0:u ( x) e ax2 c e 1a 2 4b x / 2 c2 e a 2 4b x / 2 (8.6)Case 2: a2 - 4b 0:u ( x) e ax2 1 12 2 ASin4baxBCos4ba x 22 Case 3: a2 - 4b 0:u ( x) c1 e(8.8)A special case ax2 c2 x e ax2 c1 c2 x e ax2(8.12)where c1, c2, A and B are arbitrary constants to be determined by given conditions10
Example 8.1 (p.246): Solve the following differential equation:d 2 u ( x)du ( x) 5 6u ( x) 0dxdx 2(a)Solution:We have a 5 and b 6, by comparing Equation (a) with the typical differential equation inEquation (8.1) will lead to:a2 – 4b 52 -4x6 25 – 24 1 0 - a Case 1 situation witha 2 4b 1 1Consequently, we may use the standard solution in Equation (8.6) for the general solution ofEquation (a):u ( x) eor ax2 c e 1 a 2 4b x / 2 c2 e a 2 4b x / 2 u x e 5 x / 2 c1 e x / 2 c2 e x / 2 c1 e 2 x c2 e 3 xwhere c1 and c2 are arbitrary constants to be determined by given conditions11
Example 8.2 (p.246): Solve the following differential equation with given conditions:d 2 u ( x)du ( x) 6 9u ( x) 0dxdx 2with given conditions:u(0) 2anddu ( x)dxSolution:(a)(b)(c) 0x 0Again by comparing Equation (a) with the typical differential equation in Equation (8.1), we have: a 6 andb 9. Further examining a2 – 4b 62 – 4x9 36 – 36 0, leading to special Case 3 in Equation (8.12) forthe solution:axaxax (8.12)u ( x) c e 2 c x e 2 c c x e 21or2u ( x) c1 c 2 x e6 x2 1 2 c1 c 2 x e 3 xUsing Equation (b) for Equation (d) will yield c1 2, resulting in:u(d) x 2 c2 x e 3 x(e)Differentiating Equation (e) with condition in Equation (c) will lead to the following result:du x e 3 x c2 3e 3 x 2 c2 x dx x 0 We may thus solve for c2 6Hence the complete solution of Equation (a) is: x 0 c2 6 0u ( x) 2 1 3 x e 3 x12
8.38.3.1Application of 2nd- Order Homogeneous Differential Equationsfor Free Mechanical Vibration Analysis (p.246)What is mechanical vibration and resulting consequences?Mechanical vibration is a form of oscillatory motion of a solid, a structure, a machine, ora vehicle induced by mechanical means.The amount of movement in these solids and structures is called “amplitude”. Theamplitudes of vibrating solids vary with time. Such variations may be either regularly or inrandom fashions.Oscillatory motion of solids with their amplitudes vary with fixed time interval called“period”, and the reciprocal of the period is the “frequency” of the vibratory motions.Consequences of mechanical vibrations:It can be immediate, such as in the case of resonant vibration with rapid increase ofmagnitudes of vibration, resulting in immediate and unexpected catastrophicallystructural failures, or it can induce damages accumulated by long-term vibrations withlow amplitudes. The latter form of vibrations may result in the failure of the machine orstructure due to fatigue of the materials that make the machines or structures.13
8.3.2Common Sources for Mechanical Vibrations:8.3.3Common types of Mechanical Vibrations:Amplitudes(1) Application of time-varying mechanical forces or pressure.(2) Fluid induced vibrations due to intermittent forces of wind, tidal waves, etc.(3) Application of pressures associated with acoustics and ultrasonic waves.(4) Random movements of supports, for example, seismic forces.(5) Application of thermal, magnetic forces, etc.Time, t0(1) With constant amplitudes and frequencies:AmplitudePeriod(2) With variable amplitudes but constantfrequencies:Time, t0(3) With random amplitudes and frequencies:AmplitudePeriodTime, t014
Mitigation of Mechanical Vibrations in mechanical systems:Mechanical vibrations, in the design of mechanical systems, are normally undesirableoccurrences, and engineers would normally attempt to either reduce it to the minimumappearance, or eliminate it completely.“Vibration Isolators” are commonly designed and used to minimize vibration of mechanicalsystems, such as shown in the following cases:Benches for highprecision instrumentsSuspension of heavyduty truckVibration isolatorsVibration isolatorsDesign of vibration isolators requires analyses to quantify the amplitudes and frequencies of thevibratory motion of the mechanical system – a process called “mechanical vibration analysis”15
8.3.4 Thethree types of mechanical vibration analyses by mechanical engineers (p.247):Mechanical vibration requires: Mass, spring force (elasticity), damping factor and initiator8.3.4.1 Free vibration analysis:The mechanical system (or a machine) is set to vibrate from its initial equilibrium conditionby an instantaneous disturbance (either in the form of a force or a displacement).This disturbance does not exist after the mass is set to vibrate.There are two types of free vibrations: Simple mass-spring system:MassSprings Damped vibration system:MassandSpring &Damper8.3.4.3 Forced vibration analysis (p.248):Vibration of the mechanical system is induced by cyclic loading at all times.MassForces induced bythe rugged terrainModal analysisTo identify natural frequencies of a solid machine at various possible modes of vibration16
8.4 Physical Modeling of Mechanical Vibrations: Simple mass-spring system (p. 249)Minimum requirement for Mechanical vibration: a MASS attached to an ELASTIC SUPPORTThe simplest model for mechanical vibration analysis is a MASS-SPRING system as illustrated in Figure 8.5:Massmkk MassmFigure 8.5 Simple mass-spring systemwith m mass, andk spring constant k is defined as the amount of forcerequired to deflect a certain amountof the spring F/δInduced So, k has a unit of lbf/in or N/mDeflection k is a property of a given springδ The spring in this system is to support the mass Springs in the system need not to be “coil” springs Any ELASTIC solid support can be viewed as a “spring”Simple Mass-Spring SystemsSpring:Cableor rodMassApplied forceFComplex SystemSprings:Support StructureMassSpring:Elastic beamMasses:Masses of thebridge structure17
A Case of Simple Mass-Spring Systems in Free VibrationMassmkkIt takes a MASS and SPRING (or elastic) supportto get the mass to vibrateMassmThe physical phenomena of solids in free vibration is that the vibration of the solid isinduced by an instantaneous disturbance either in the form of a force or deformationof the supporting spring, such as illustrated in the vibration of a vehicle induced by itssuspension system:.MassVehicle mass per WheelmThis initial disturbance does notexist after the inception ofSuspension of the vehiclevibration of the solid mass.Motion ofThe vehicleSpring Constant, kThis disturbance (a “blip” onRoad surface) causes the mass to begin vibration,and continue to vibrate afterward18
8.4 Mathematical Formulation of Simple Mass-Spring Systems in Free Vibration (p.249)MassmAs we mentioned in previous occasions: “The simplest physical modelfor mechanical vibration analysis involves a MASS and a SPRING(or an elastic support) as illustrated in Figure 8.5kkMassmMathematical Formulation for Free Vibration of a mass:kkkInitial equilibrium position(1) We will Begins with:hMassm(a) Free-hung spring(2) The free-hung spring deflectsupon attaching a mass m:Stretching of Springat t is:h y(t)-y(t)MassmInitial equilibriumpositionwith a mass attime zero y(t) Sign convention: ve downward(b) Statically stretchedspring(c) A vibrating mass at time t(3) A small instantaneous “push-down” is applied to the massand release quickly.We can expect the mass to bounce up & down passingits initial equilibrium position. Due to simultaneous effects ofthe recoil of the spring and the dynamic forces associatedwith the motion of the mass in variable velocities.19
Mathematical formulation of a vibrating massin a Mass-Spring SystemkkkMassmh(a) Free-hung springStatic force(b) StaticallystretchedEquilibriumspringStretching of Springat t is:h y(t)mMassm-y(t) y(t)Based on an assumption with no air resistance againstthe motion off the mass and spring. Formulation is onthe equilibrium of of both static and dynamic forces.Forces: Weight (W); Spring force (Fs)Dynamic force, F(t):Spring force:Dynamic (Inertia)Fs k[h y(t)]Force, F(t)Displacement y(t):Position at time tm y(c) ADynamicvibrating massat time tforceeEquilibrium at time tSpring forceFs kh Fy W Fs 0mmg khWeightW mgmWeightW mgEquilibrium of forces acting on the mass at given time tsatisfies the Newton’s 1st Law: F t Fs W 0d 2 y(t )But since we have the dynamic force to be: F (t ) mdt 2and the spring force to be Fs k[h y(t)], we should have:d 2 y (t ) m k h y (t ) mg 0dt 2But mg kh from the static equilibrium condition,and after substituting it into the above equation, we have thefollowing 2nd order differential equation for the instantaneous position y(t) for the vibrating mass:d 2 y (t )m k y (t ) 0dt 2(8.14)20
Solution of differential equation (8.14) for simple mass-spring vibrationyd 2 y (t )m k y (t ) 0dt 2y(t)Massm(8.14)kwhere y(t) instantaneous position of the masskMassmRe-writing the above equation in the following form:y(t)yd 2 y t k y t 0dt 2m(8.14a)The solution of Equation (8.14) can be obtained by comparing Equation (8.14a) with the typical 2ndorder differential equation in Equation (8.1):du( x)d 2u ( x )(8.1) bu( x) 0a2dxdxWe will find that a 0 and b k/m after comparing Eqs.(8.14a) and (8.1). The solution of Equation (8.1)depends on the discriminator: a2 – 4b. Since k spring constant which is a property of the spring andm mass of the vibrating solid, the equivalent coefficient b is a ve real number. Consequently, we willhave: a2 – 4b 0 – 4(k/m) 0, which is a Case 2 for the solution, as shown in Equation(8.8), ory (t ) A Coskkt B Sintmm(8.15)where A, B are arbitrary constants to be determined by given conditions21
Physical senses of solution of differential equation (8.14) for simple mass-spring vibration:The differential equation for the instantaneous position of the solid mass, y(t) satisfies thefollowing Equation (8.14):y(8.14)d 2 y (t )m k y (t ) 0y(t)dt 2where y(t) instantaneous position of the massMassmkUpon re-writing the equation in the form:kd 2 y t k y t 0dt 2mMassmy(t)y(8.14a)The general solution of Equation (8.14) is:y (t ) c1 Cos o t c2 Sin o t(8.16)where c1 and c2 are arbitrary constants to be determined by given conditions, and o km(8.16a)is called the “circular”, or “angular frequency” of the mass-spring vibration system. Often, it represents the“natural frequency” of the simple mass-spring system. The unit is rad/s.Corresponding to the angular frequency ωo is the real frequency of the vibration in the following expression::f o1 2 2 km(8.17)22
Graphical representation of free-vibration of simple mass-spring systems (p.252):The mathematical solution for the instantaneous position of the vibrating mass can be obtained bythe following expression:y (t ) c1 Cos o t c2 Sin o ty(8.16)The solution y(t) consists of cosine and sine functionsof variable t (the time)y(t)MassmSo, it is an oscillatory function, oscillating about the“zero-time” axis, with the amplitudes of vibration y(t)illustrated in Figure 8.8:kkMassmAmplitude, y(t)y(t)y(0) Initialdeflection ofthe springyCombination of cosine and sine functionsMax. amplitudeTime, t0Period 2 kmThe frequency of vibration: f 1/periodFigure 8.8 The oscillatory motion of the solid mass in a simple mass-spring system23
Example 8.3 (p.253): An unexpected case for engineers to consider in their design and operation ofan unloading process of a heavy machine.Description of the problem:A truck is unloading a heavy machine weighingElastic cable with800 lbf by a crane as illustrated in Figure 8.10.k 6000 lbf/inv 20 ft/min The cable used to lift the freight was suddenlyseized (jammed) at time t from a descendingvelocity of v 20 ft/min800 lbfOne may expect the heavy machine undergoingAn “up-down-up” vibration after such seizure.Figure 8.10Determine the following:(A) The frequency of vibration of the machine that is seized from its descending(B) The maximum tension in the cable induced by the vibrating machine,(C) The maximum stress in the cable if the stranded steel cable is 0.5 inch in diameter(D) Would the cable break if its ultimate tensile strength (UTS) of the cable material is 40,000 psi?Solution:Because the machine is attached to anelastic cable, which has the same characteristicsas a “spring,” we may simulate this situationto a simple mass-spring systems as:ElasticCableSpringconstantk 6000 lb/inMass:800/32.2slugSpringThe frequency and amplitudes of the vibrating machine can thus be evaluated by the expressionsderived for the simple mass-spring system.24
Example 8.3 – Cont’d(a) The frequency of vibration of the machine is given in Equation (8.17).The circular frequency is: o k m6000 x12 53.83 rad / s800 / 32.2which leads to the frequency to be:f o 8.57 cycles / s2 (b) The maximum tension in the cable:The maximum tension in th
8.2 Typical form of second-order homogeneous differential equations (p.243) ( ) 0 2 2 bu x dx du x a d u x (8.1) where a and b are constants The solution of Equation (8.1) u(x) may be obtained by ASSUMING: u(x) emx (8.2) in which m is a constant to be determined by the following procedure: If the assumed solution u(x) in Equation (8.2) is a valid solution, it must SATISFY
Part One: Heir of Ash Chapter 1 Chapter 2 Chapter 3 Chapter 4 Chapter 5 Chapter 6 Chapter 7 Chapter 8 Chapter 9 Chapter 10 Chapter 11 Chapter 12 Chapter 13 Chapter 14 Chapter 15 Chapter 16 Chapter 17 Chapter 18 Chapter 19 Chapter 20 Chapter 21 Chapter 22 Chapter 23 Chapter 24 Chapter 25 Chapter 26 Chapter 27 Chapter 28 Chapter 29 Chapter 30 .
TO KILL A MOCKINGBIRD. Contents Dedication Epigraph Part One Chapter 1 Chapter 2 Chapter 3 Chapter 4 Chapter 5 Chapter 6 Chapter 7 Chapter 8 Chapter 9 Chapter 10 Chapter 11 Part Two Chapter 12 Chapter 13 Chapter 14 Chapter 15 Chapter 16 Chapter 17 Chapter 18. Chapter 19 Chapter 20 Chapter 21 Chapter 22 Chapter 23 Chapter 24 Chapter 25 Chapter 26
DEDICATION PART ONE Chapter 1 Chapter 2 Chapter 3 Chapter 4 Chapter 5 Chapter 6 Chapter 7 Chapter 8 Chapter 9 Chapter 10 Chapter 11 PART TWO Chapter 12 Chapter 13 Chapter 14 Chapter 15 Chapter 16 Chapter 17 Chapter 18 Chapter 19 Chapter 20 Chapter 21 Chapter 22 Chapter 23 .
About the husband’s secret. Dedication Epigraph Pandora Monday Chapter One Chapter Two Chapter Three Chapter Four Chapter Five Tuesday Chapter Six Chapter Seven. Chapter Eight Chapter Nine Chapter Ten Chapter Eleven Chapter Twelve Chapter Thirteen Chapter Fourteen Chapter Fifteen Chapter Sixteen Chapter Seventeen Chapter Eighteen
18.4 35 18.5 35 I Solutions to Applying the Concepts Questions II Answers to End-of-chapter Conceptual Questions Chapter 1 37 Chapter 2 38 Chapter 3 39 Chapter 4 40 Chapter 5 43 Chapter 6 45 Chapter 7 46 Chapter 8 47 Chapter 9 50 Chapter 10 52 Chapter 11 55 Chapter 12 56 Chapter 13 57 Chapter 14 61 Chapter 15 62 Chapter 16 63 Chapter 17 65 .
HUNTER. Special thanks to Kate Cary. Contents Cover Title Page Prologue Chapter 1 Chapter 2 Chapter 3 Chapter 4 Chapter 5 Chapter 6 Chapter 7 Chapter 8 Chapter 9 Chapter 10 Chapter 11 Chapter 12 Chapter 13 Chapter 14 Chapter 15 Chapter 16 Chapter 17 Chapter
Chapter 3 Chapter 4 Chapter 5 Chapter 6 Chapter 7 Chapter 8 Chapter 9 Chapter 10 Chapter 11 Chapter 12 Chapter 13 Chapter 14 Chapter 15 Chapter 16 Chapter 17 Chapter 18 Chapter 19 Chapter 20 . Within was a room as familiar to her as her home back in Oparium. A large desk was situated i
The Hunger Games Book 2 Suzanne Collins Table of Contents PART 1 – THE SPARK Chapter 1 Chapter 2 Chapter 3 Chapter 4 Chapter 5 Chapter 6 Chapter 7 Chapter 8. Chapter 9 PART 2 – THE QUELL Chapter 10 Chapter 11 Chapter 12 Chapter 13 Chapter 14 Chapter 15 Chapter 16 Chapter 17 Chapt
